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Confusing proof made less confusing

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aisejohan committed Sep 3, 2019
1 parent f6c9e4e commit 21543f65e829e87052175a5ef25733f3af116fb1
Showing with 9 additions and 13 deletions.
  1. +9 −13 algebra.tex
The case $n = 1$. We may replace $R$ by its integral closure in $S$
(Lemma \ref{lemma-quasi-finite-permanence} guarantees that $R \to S$
is still quasi-finite at $\mathfrak q$). Thus we may assume
$R \subset S$ is integrally closed in $S$.
$R \subset S$ is integrally closed in $S$, in other words $R = S'$.
Consider the map $\varphi : R[x] \to S$, $x \mapsto x_1$.
(We will see that $\varphi$ is not injective below.)
By assumption $\varphi$ is finite. Hence we are in Situation
Thus there exists an element $s \in J$,
$s \not\in \mathfrak q$. By definition of $J$ we may write
$s = \varphi(f)$ for some polynomial $f \in R[x]$.
Now let $I = \Ker(R[x] \to S)$. Since $\varphi(f) \in J$
Let $I = \Ker(\varphi : R[x] \to S)$. Since $\varphi(f) \in J$
we get $(R[x]/I)_f \cong S_{\varphi(f)}$. Also $s \not \in \mathfrak q$
means that $f \not \in \varphi^{-1}(\mathfrak q)$. Thus
$\varphi^{-1}(\mathfrak q)$ is a prime of $R[x]/I$
at which $R \to R[x]/I$ is quasi-finite, see
Lemma \ref{lemma-quasi-finite-local}.
Let $C \subset R[x]/I$ be the integral closure of $R$. By
Lemma \ref{lemma-quasi-finite-local}. Note that $R$ is integrally closed
in $R[x]/I$ since $R$ is integrally closed in $S$. By
Lemma \ref{lemma-quasi-finite-monogenic}
there exists an element $h \in C$, $h \not \in \varphi^{-1}(\mathfrak q)$
such that $C_h \cong (R[x]/I)_h$. We conclude that
there exists an element $h \in R$, $h \not \in R \cap \mathfrak q$
such that $R_h \cong (R[x]/I)_h$. Thus
$(R[x]/I)_{fh} = S_{\varphi(fh)}$ is isomorphic to a principal
localization $C_{h'}$ of $C$ for some
$h' \in C$, $h' \not \in \varphi^{-1}(\mathfrak q)$.
Since $\varphi(C) \subset S'$ we get
$g = \varphi(h') \in S'$, $g \not \in \mathfrak q$
and moreover the injective map $S'_g \to S_g$ is also surjective
because by our choice of $h'$ the map $C_{h'} \to S_g$ is surjective.
localization $R_{h'}$ of $R$ for some
$h' \in R$, $h' \not \in \mathfrak q$.

\medskip\noindent
The case $n > 1$. Consider the subring $R' \subset S$
which is the integral closure of $R[x_1, \ldots, x_{n-1}]$
in $S$. By Lemma \ref{lemma-four-rings} the extension
in $S$. By Lemma \ref{lemma-quasi-finite-permanence} the extension
$S/R'$ is quasi-finite at $\mathfrak q$. Also, note
that $S$ is finite over $R'[x_n]$.
By the case $n = 1$ above, there exists a $g' \in R'$,

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